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Prove that the center of $S_n$ is the identity subgroup for $n>2$.

Would I have to use the definition of center of a group first or do I have to make sure it satisfies the axioms to be a subgroup? I am not too sure what the proof wants me to show. I am not sure how to go about solving it.

Mike Pierce
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Sam
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  • @VikrantDesai i want to show this whole center S_n is the identity subgroup not that it just contains an element that is the identity – Sam Feb 11 '17 at 04:47
  • Read that link properly. Every group has an identity. Problem is to show that there is no non-identity element in the center of $S_n (n \gt 2)$. Go through both answers in that link. – Error 404 Feb 11 '17 at 04:50
  • @VikrantDesai so the issue would arise if there were no identity element? I looked through the answers, so these are a proof by contradiction? – Sam Feb 11 '17 at 04:56
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    @Sam you need to know the definition of the center (all elements that commute with every other element). From that you can prove that the center is always a subgroup but that's not the issue at hand. For the problem, you need to show that the center of $S_n$ only contains one element: the identity. In other words that there are no permutations in $S_n$ that compute with every other permutation. Mike Pierce's answer below looks like a good way to do that. – spaceisdarkgreen Feb 11 '17 at 05:02
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    The center of a group is always a subgroup. You don't have to check that separately for this exercise, but if you've never verified that before, you should. And yeah, if you are proving something about the center of a group, the definition of the center will need to come up somewhere in your proof (you should be concerned if it doesn't). – Mike Pierce Feb 11 '17 at 05:08
  • @MikePierce okay thank you. I will look over all this closely as I work on this problem. I am only just learning all this so I was confused, but these explanations are making more sense. – Sam Feb 11 '17 at 05:12

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We need to show that the center of a group, $$\operatorname{C}(G) = \left\{x \in G \mid axa^{-1} = x\;\;\forall a \in G\right\}\,,$$ contain the identity element, and only the identity element. It's easy to show that the identity element is in ${\operatorname{C}(G).}$

Now suppose (for the sake of contradiction) that we have a non-identity element $\sigma {\in \operatorname{C}(G).}$ To keep things simple for now, suppose $\sigma$ is a cycle $(x_1\;\dotsb\;x_m)$. Recall that (or you should prove that) $$\pi(x_1\;\dotsb\;x_m)\pi^{-1} = (\pi(x_1)\;\dotsb\;\pi(x_m))\,.$$ Knowing this, for any given $\sigma$ do you think you can find a $\pi$ that won't fix $\sigma$ under conjugation? Then can you see how this argument extends to elements of $S_n$ that are not cycles?

If there is a element $i \in \{1,\dotsc,n\}$ such that $x_i$ is not moved by the permutation $\sigma$, then let $\pi = (x_i,x_1)$. Otherwise $(x_1\;\dotsb\;x_m)$ permutes at least three elements, and you may let $\pi = (x_1,x_m)$. Since every permutation of $S_n$ can be written as a product of disjoint cycles, this argument on a single cycle $\sigma$ extends to any permutation in $S_n$.

Mike Pierce
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